Posts Tagged ‘the sun’s gravitational force’

Last time we discovered that Earth zips around the sun at the mind boggling speed of 29,680 meters per second. This is the final bit of information required to calculate Fg, the gravitational force exerted upon Earth by its sun, as set out in Newton’s equation on the subject and derived from his Second Law of Motion. We’ll calculate that quantity today.

This metric unit of force, kilogram • meter per second2, represents kilograms multiplied by meters, and their product divided by seconds squared. It’s known in scientific circles as the Newton, in honor of Sir Isaac Newton, widely recognized as one of the greatest scientists of all time and a key figure in the scientific revolution that began over three centuries ago. Therefore the sun’s gravitational force acting upon Earth is typically referred to as,

Fg = 3.52 × 1022Newtons

Here in the US where we like to use English units such as feet and pounds, the Newton is said to equal 0.225 pounds of force. Therefore in English units the sun’s gravitational force is expressed as,

Fg = (3.52 × 1022Newtons) × (0.225 pounds of force per Newton)

Fg = 7.93 × 1021 pounds

That’s scientific notation for 7,930,000,000,000,000,000,000 pounds! That’s the amount of force exerted by the sun’s gravitational pull on Earth. Seems about right — right?

Now that we know Fg, we have everything we need to calculate the mass of the sun, which in turn enables us to determine the mass and gravity of other planets in our solar system. We’ll calculate the sun’s mass next time.